ChatterBank1 min ago
Photon absorption, radiation and frequencies
1 Answers
Hello, these questions are interrelated. Thanks
Q1
I read that absorption of a photon will occur only when the quantum energy of the photon precisely matches the energy gap between the initial and final states.
The converse is also true and a photon of the same frequency is emitted when the system drops from the upper to the lower state.
In other words, only photons of specific frequencies can be absorbed or emitted from a body.
I understand that the Planck curve of emitted photon frequencies from a body is constant for a given temperature and would appear to cover a very large number of frequencies and the frequencies are not dependant on the nature of the emitter . Thus at a given temperature every body emits the same frequencies and energy levels.
How can these two apparently conflicting concepts be reconciled?
Q2
If a black body 'oven' is constructed from copper, and is at 300K. We know that radiant energy will be emitted from the copper as per Planck's curve.
What puzzles me is what happens to all those photons that cannot be re-adsorbed because they don't have the correct frequencies?
If they were to through the wall of the 'oven' and out of the system then the inside of the 'oven' would be losing energy and get colder.
This doesn't happen, so what is going on here?
Q3
Assume that I have a photon generator that can beam great quantities of photons at a fixed temperature (Planck curve) at a target body.
How will the photons impart their energy to the target body? What will happen when the body reaches the same temperature as the photons? The second law of thermodynamics indicates that the target temperature will not rise beyond the temperature of the photons. I am trying to understand the physical process that causes this.
Q1
I read that absorption of a photon will occur only when the quantum energy of the photon precisely matches the energy gap between the initial and final states.
The converse is also true and a photon of the same frequency is emitted when the system drops from the upper to the lower state.
In other words, only photons of specific frequencies can be absorbed or emitted from a body.
I understand that the Planck curve of emitted photon frequencies from a body is constant for a given temperature and would appear to cover a very large number of frequencies and the frequencies are not dependant on the nature of the emitter . Thus at a given temperature every body emits the same frequencies and energy levels.
How can these two apparently conflicting concepts be reconciled?
Q2
If a black body 'oven' is constructed from copper, and is at 300K. We know that radiant energy will be emitted from the copper as per Planck's curve.
What puzzles me is what happens to all those photons that cannot be re-adsorbed because they don't have the correct frequencies?
If they were to through the wall of the 'oven' and out of the system then the inside of the 'oven' would be losing energy and get colder.
This doesn't happen, so what is going on here?
Q3
Assume that I have a photon generator that can beam great quantities of photons at a fixed temperature (Planck curve) at a target body.
How will the photons impart their energy to the target body? What will happen when the body reaches the same temperature as the photons? The second law of thermodynamics indicates that the target temperature will not rise beyond the temperature of the photons. I am trying to understand the physical process that causes this.
Answers
Best Answer
No best answer has yet been selected by hapanabasa. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.An atom can absorb more than 1 photon and can move up into higher energy states. It can jump over some energy states and go to higher energies directly. Similarly when emitting photons it can drop down into lower energies by skipping intermediate states. This means that it can emit or absorb photons of different frequencies/energies, not just one.
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